For the decompostion of the compound- represented as NH2COONH4s- 2NH3g-CO2g- the Kp-2-9 - 10-5 atm3- If the reaction is started with 1 mol of the compounds- the total pressure at equilibrium woud be-

The correct option is C 5.80 × 10^ 2 atm NH_4OCONH_2 (s) ⇋ 2NH_3 (g) + CO_2 (g) #160; #160; #160; #160; #160; #160; #160; #160 ...

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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

For the decom...

Question

For the decompostion of the compound, represented as $N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)$ ; the $K_{p} = 2.9 \times 10^{- 5} a t m^{3}$ . If the reaction is started with 1 mol of the compounds, the total pressure at equilibrium woud be:

1.94 \times 10^{- 2} a t m

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5.80 \times 10^{- 2} a t m

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7.66 \times 10^{- 2} a t m

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38.8 \times 10^{- 2} a t m

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Solution

The correct option is C $5.80 \times 10^{- 2} a t m$
$N H_{4} O C O N H_{2} (s) ⇋ 2 N H_{3} (g) + C O_{2}$ (g)
2x x

$K_{p} = (P_{N H_{3}})^{2} \times (P_{C O_{2}})$

$2.9 \times 10^{- 5} = 2^{2} \times x^{3}$

$x = 0.019 a t m =$ carbon dioxide pressure

So the pressure for $N H_{3}$ is twice of that according to chemical equation.

Total pressure $= (2 \times 0.0193 + 0.0193) a t m$

Total pressure $= 5.8 \times 10^{- 2} a t m$

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Similar questions

Q. For the decomposition of the compound, represented as:

N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)

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K_{P} = 2.9 \times 10^{- 5} a t m^{3}

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When 1.0 mole of $N_{2}$ and 3.0 moles of $H_{2}$ was heated in a vessel at 773K and a pressure of 3.55 atm. 30% of $N_{2}$ is converted into $N H_{3}$ at equilibrium. Calculate KP for the reaction -

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$N H_{2} C O O N H_{4} (s) ⇋ 2 N H_{3} (g) + C O_{2} (g)$ For the reaction, $K_{P} = 2.9 \times 10^{- 5} a t m^{3}$ . Calculate the total pressure at equilibrium if $1 m o l e o f N H_{2} C O O N H_{4}$ is taken.

Q. Ammonium carbonate decomposes as

{N H}_{2} {C O O N H}_{4 (s)} ⇌ {2 N H}_{3 (g)} + {C O}_{2 (g)}

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Q. For the decomposition,

N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)

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K_{p} = 2.9 \times 10^{- 5} a t m^{3}

The total pressure of gases at equilibrium when 1.0 mol of

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For the decompostion of the compound, represented as NH2COONH4(s)⇌2NH3(g)+CO2(g); the Kp=2.9×10−5atm3. If the reaction is started with 1 mol of the compounds, the total pressure at equilibrium woud be:

The correct option is C 5.80×10−2atmNH4OCONH2(s)⇋2NH3(g)+CO2 (g) 2x xKp=(PNH3)2×(PCO2)2.9×10−5=22×x3 x=0.019atm= carbon dioxide pressure So the pressure for NH3 is twice of that according to chemical equation. Total pressure =(2×0.0193+0.0193) atmTotal pressure =5.8×10−2 atm

For the decompostion of the compound, represented as $N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)$ ; the $K_{p} = 2.9 \times 10^{- 5} a t m^{3}$ . If the reaction is started with 1 mol of the compounds, the total pressure at equilibrium woud be: